Graphical Method — Drawing the Line of a Linear Equation
Every linear equation in two variables represents a straight line on the Cartesian plane. Exercise 6.3 focuses entirely on this graphical representation — you calculate two or three solution points, plot them, and join them to draw the line. The line then reveals values graphically that would otherwise require algebra.
Steps to draw the graph of a linear equation:
Step 1: Rearrange the equation into the form y = f(x).
Step 2: Choose convenient values of x and compute the corresponding y values.
Step 3: Write the ordered pairs (x, y).
Step 4: Plot the points on graph paper.
Step 5: Join the points with a straight line and extend it in both directions.
Every point on this line is a solution of the equation. Two points are enough to draw the line, but three are recommended to catch errors.
Step 1: Rearrange the equation into the form y = f(x).
Step 2: Choose convenient values of x and compute the corresponding y values.
Step 3: Write the ordered pairs (x, y).
Step 4: Plot the points on graph paper.
Step 5: Join the points with a straight line and extend it in both directions.
Every point on this line is a solution of the equation. Two points are enough to draw the line, but three are recommended to catch errors.
Worked Example — Graph of 3x + y = 7
Rearrange: y = 7 − 3x
x = 1 → y = 7 − 3 = 4 → Point: (1, 4)
x = 3 → y = 7 − 9 = −2 → Point: (3, −2)
x = 0 → y = 7 − 0 = 7 → Point: (0, 7)
Plot (1,4), (3,−2), (0,7) and join → straight line for 3x + y = 7.
Exercise 6.3 — Question 1: Draw the Graph of Four Equations
(i) 2y = −x + 1
Solution Table — 2y = −x + 1 ⟹ y = (−x + 1) / 2
| x | y = (−x+1)/2 | (x, y) |
|---|---|---|
| 1 | 0 | (1, 0) |
| 3 | −1 | (3, −1) |
| 5 | −2 | (5, −2) |
Figure 1(i): Graph of 2y = −x + 1 — a line with negative slope passing through (1, 0).
(ii) −x + y = 6
Solution Table — −x + y = 6 ⟹ y = 6 + x
| x | y = 6 + x | (x, y) |
|---|---|---|
| −3 | 3 | (−3, 3) |
| 0 | 6 | (0, 6) |
| 2 | 8 | (2, 8) |
The y-intercept is (0, 6). The line has slope +1 and rises left to right. All three points align on the same straight line confirming the graph.
(iii) 3x + 5y = 15
Solution Table — 3x + 5y = 15 ⟹ y = (15 − 3x) / 5
| x | y = (15−3x)/5 | (x, y) |
|---|---|---|
| 0 | 3 | (0, 3) |
| 5 | 0 | (5, 0) |
| −5 | 6 | (−5, 6) |
x-intercept: (5, 0). y-intercept: (0, 3). The line cuts both axes — useful reference points for graphing.
(iv) x/2 − y/3 = 3
Solution Table — x/2 − y/3 = 3 ⟹ y = (3x/2 − 9) or y = 3(x/2 − 3)
| x | y | (x, y) |
|---|---|---|
| 0 | −9 | (0, −9) |
| 6 | 0 | (6, 0) |
| 4 | −3 | (4, −3) |
x-intercept: (6, 0). y-intercept: (0, −9). This line has a steep positive slope and starts well below the x-axis on the y-axis side.
Exercise 6.3 — Question 2: Equations of the Form y = mx
Five equations are graphed together: y = x, y = 2x, y = −2x, y = 3x, y = −3x. All are in the form y = mx.
y = x
(1,1), (2,2), (3,3)
Slope = 1, passes through origin
y = 2x
(2,4), (3,6), (4,8)
Slope = 2, steeper than y = x
y = −2x
(2,−4), (3,−6), (4,−8)
Slope = −2, falls left to right
y = 3x
(−2,−6), (3,9), (−4,−12)
Slope = 3, steepest positive
y = −3x
(−2,6), (3,−9), (−4,12)
Slope = −3, steepest negative
(i) Yes — all five equations are of the form y = mx where m is a real number.
(ii) Yes — all five graphs pass through the origin (0, 0).
(iii) Conclusion: All lines of the form y = mx pass through the origin. The value of m (the slope) determines how steep the line is and whether it rises (m > 0) or falls (m < 0) from left to right.
(ii) Yes — all five graphs pass through the origin (0, 0).
(iii) Conclusion: All lines of the form y = mx pass through the origin. The value of m (the slope) determines how steep the line is and whether it rises (m > 0) or falls (m < 0) from left to right.
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Exercise 6.3 — Questions 3 to 5: Read Values from the Graph
Question 3 — 2x + 3y = 11, find y when x = 1
Solution — 2x + 3y = 11 ⟹ y = (11 − 2x) / 3
x = −2 → y = (11+4)/3 = 5 → (−2, 5)
x = 4 → y = (11−8)/3 = 1 → (4, 1)
x = −5 → y = (11+10)/3 = 7 → (−5, 7)
From the graph: draw vertical line at x = 1, read y-value where it meets the line
When x = 1, y = 3 (read from graph)
Question 4 — y − x = 2, find y when x = 4 and x when y = −3
Solution — y − x = 2 ⟹ y = 2 + x
x = 1 → y = 3 → (1, 3)
x = 3 → y = 5 → (3, 5)
x = −1 → y = 1 → (−1, 1)
(i) At x = 4: y = 2 + 4 = 6 → confirmed from graph
(ii) At y = −3: x = −3 − 2 = −5 → confirmed from graph
(i) y = 6 when x = 4. (ii) x = −5 when y = −3
Question 5 — 2x + 3y = 12, find solutions from graph
Solution — 2x + 3y = 12 ⟹ y = (12 − 2x) / 3
x = 3 → y = 2 → (3, 2)
x = 6 → y = 0 → (6, 0)
x = 0 → y = 4 → (0, 4)
(i) When y = 3: 2x + 9 = 12 → x = 1.5 → Solution: (1.5, 3)
(ii) When x = −3: −6 + 3y = 12 → y = 6 → Solution: (−3, 6)
(i) Point with y = 3 is (1.5, 3). (ii) Point with x = −3 is (−3, 6)
Exercise 6.3 — Question 6: Find Axis-Intercepts from Graph
For each equation, draw the graph and identify where the line crosses the x-axis (y = 0) and the y-axis (x = 0). These intercept points are particularly easy to calculate.
(i) 6x − 3y = 12
⟹ y = 2(x − 2)
Points: (3,2), (5,6), (1,−2)
x-intercept: set y=0 → x=2
y-intercept: set x=0 → y=−4
Cuts x-axis at (2,0), y-axis at (0,−4)
(ii) −x + 4y = 8
⟹ y = (8 + x) / 4
Points: (4,3), (−4,1), (8,4)
x-intercept: set y=0 → x=−8
y-intercept: set x=0 → y=2
Cuts x-axis at (−8,0), y-axis at (0,2)
(iii) 3x + 2y + 6 = 0
⟹ y = (−3x − 6) / 2
Points: (2,−6), (4,−9), (−4,3)
x-intercept: set y=0 → x=−2
y-intercept: set x=0 → y=−3
Cuts x-axis at (−2,0), y-axis at (0,−3)
Exercise 6.3 — Questions 7–10: Real-Life Graph Applications
Question 7 — Rajiya and Preethi collected ₹1000 for PM Relief Fund
Equation Setup
Let Rajiya's contribution = ₹x, Preethi's = ₹y
Equation: x + y = 1000 ⟹ y = 1000 − x
Points: (200, 800), (400, 600), (600, 400)
Graph is a straight line with x-intercept (1000, 0) and y-intercept (0, 1000).
Question 8 — Gopaiah's wheat and paddy fields, total area 5000 sq.m.
Equation Setup
Let area under wheat = x sq.m., area under paddy = y sq.m.
Equation: x + y = 5000 ⟹ y = 5000 − x
Points: (1000, 4000), (3000, 2000), (4000, 1000)
Same form as Q7 — a straight line cutting both axes at 5000.
Question 9 — Force on a 6 kg body: F = ma
Equation Setup — Newton's Second Law
Mass m = 6 kg. Force F is directly proportional to acceleration a.
F ∝ a ⟹ F = ma ⟹ F = 6a
Points (a, F): (1, 6), (3, 18), (5, 30)
Graph is a straight line through the origin — y = 6x type (passes through origin). Slope = 6 = mass of the body.
Question 10 — Falling stone: V = 9.8t, find V at t = 4 sec
Equation Setup
V = 9.8t (velocity in m/s, time in seconds)
Points (t, V): (5, 49), (10, 98)
At t = 4: V = 9.8 × 4 = 39.2 ≈ 39 m/s
From the graph, at t = 4 seconds, velocity ≈ 39 m/s (approximately).
Key Rules — Graphical Method Summary
- Every linear equation in two variables graphs as a straight line — you only need two points, but always plot three to verify.
- To find the x-intercept, set y = 0 and solve for x. To find the y-intercept, set x = 0 and solve for y.
- All equations of the form y = mx pass through the origin (0, 0). The slope m controls steepness and direction.
- To read a value from the graph: draw a vertical line at the known x-value, find where it meets the line, then read the y-value horizontally.
- Real-life applications — distance-time, force-acceleration, cost-quantity — all produce linear equations whose graphs give instant answers for any input value.